Question

The inductance of attraction type instrument is given by L = (10 + 5θ − θ2)μH where θ is the deflection in radian from zero position. The spring constant is 12 × 10−6 N-m/rad. Find out the deflection for a current of 5A.

Answer 1

L = (10 + 5θ − θ2) × 10−6 H

∴ dL/dθ = (0 + 5 − 2 × θ) × 10–6

= (5 − 2θ) × 10−6 H/rad

Let the deflection be θ radians for a current of 5A, then deflecting torque,

Td = 12 × 10−6 × θ N-m

Also, Td = 1/2 I2 dL/dθ

Equating the two torques, we get

12 × 10−6 × θ = 1/2 × 52 × (5 − 2θ) × 10−6

∴ θ = 1.689 radian